Orbits of the dynamical system $\dot r=-r\log r$, $\dot\varphi=2r\sin^2\left(\varphi/2\right)$

Question

I have a question regarding this specific 2-dimensional DE: $$\dot r=-r\log r\\\dot\varphi=2r\sin^2\left(\frac{\varphi}{2}\right)$$ where $r> 0$ and $\varphi\in [0,\infty)$ (yes, $2\pi$ would do it as well).

Now, I want to see how the stationary point $(1,0)$ is attractive. It is easy to see that $\dot\varphi>0$ at all times $t$ so all solutions constantly rotate until hitting a stationary point. For starting values left of $(1,0)$ (so $r<1$) we also see that $\dot r>0$ and for starting values right of $(1,0)$ we have $\dot r<0$. So let's say we start at some point in the first quadrant left-above of $(1,0)$ (for example $\frac{1}{2}(1,1)$). Then the solution moves into the direction of the unit circle while also doing a spiral since $\dot\varphi >0$. But how does this work? How can $\varphi$ and $r$ both increase at all times when I start left-above the stationary point $(1,0)$? How does it ever hit $(1,0)$ without decreasing again?

I hope, I could make my problem clear enough. If not, I can provide a sketch of what I mean. Maybe also my question can be answered if someone provides a sketch of the trajectory starting at the mentioned point $(\frac{1}{2},\frac{1}{2})$.

Answer 1

If $0<r<1$, we have $\dot{r}>0$, which means $r$ will increase. If $1<r$, then $\dot{r}<0$, so $r$ will decrease. This implies that $r=1$ is attractive, period. No matter where the system starts (other than at the undefined origin), $r=1$ is attractive.

The exact solution of the $r$ equation is $r(t)=e^{e^{-t+C}}$ or $r(t)=e^{A\,e^{-t}},$ where $A=e^C$. This is always positive.

Turning our attention to the $\varphi$ equation, we see that $\dot{\varphi}\ge 0$ for all $r,\varphi$ (and hence for all time). The stationary points for $\varphi$ are: $$\frac{\varphi}{2}=n\pi\quad\implies\quad \varphi=2n\pi.$$ Therefore, we say that wherever $\varphi$ starts, it will increase until the next integer multiple of $2\pi$, which it will approach asymptotically without exceeding. In the polar system, therefore, there is really only one unique attractor point: $(1,0)$.

Answer 2

Hint

From

$$ \dot r=-r\log r\rightarrow -\frac{dr}{r\log r} = dt $$

or

$$ \log r = C_0 e^{-t}\to r = e^{C_0 e^{-t}} $$

or also

$$ \frac{dr}{d\varphi} = -\frac{\log r}{2\sin^2(\frac{\varphi}{2})} $$

This is a separable DE and can be arranged as

$$ -\frac{dr}{\log r} = \frac{d\varphi}{2\sin^2(\frac{\varphi}{2})} $$

etc.

After a convenient change of variables,

$$ x = r\cos\varphi\\ y = r\sin\varphi\\ \dot x = \cos\varphi \dot r - r\sin\varphi \dot\varphi\\ \dot y = \sin\varphi \dot r + r\cos\varphi \dot \varphi $$

and

$$ \dot r = -\sqrt{x^2+y^2}\log(\sqrt{x^2+y^2})\\ \dot \varphi = 2\sqrt{x^2+y^2}\sin^2\left(\frac{\tan^{-1}(\frac yx)}{2}\right) $$

and after a little elaboration we have the stream plot